Prove that if $\inf\{b_n\} = 0$, then $\inf\{a_n\} = 0$

Question

Suppose that $\forall n\in \mathbb N\exists a_n, b_n\in\mathbb R$ s.t. $0 < a_n < b_n$.

Prove that

If inf$\{b_n\} = 0\quad \Rightarrow\quad $ inf$\{a_n\} = 0$

So we got to prove two parts.

i) $0 \leq a_n\;\forall a_n\in \mathbb{R}$

ii) If $b \leq a_n\forall a_n\in \mathbb{R}$ then $0 \leq b$

My try:

i) We know by hypothesis that $0 < a_n \forall a_n\in\mathbb{R}$

ii) If $b \leq a_n \forall a_n\in \mathbb{R}$

then $0 < b \leq a_n < b_n$

then $0 < b \leq$ Inf$\{a_n\} < $ Inf$\{b_n\} = 0 $

then $0 \leq b$

I don't know if it's correct.

Answer 1

Part (i) looks good. For part (ii), your general idea is right, but you have a small error in your definition of what you have to show:

ii) If $b \leq a_n\forall a_n\in \mathbb{R}$ then $0 \leq b$

Actually you should show that if $b \le a_n$ for all $a_n \in \mathbb{R}$, then $b \le 0$. (So $b$ could be negative). Remember that infimum = "greatest lower bound". So you want to show that for any lower bound $b$ of $a_n$, that lower bound is less than or equal to $0$.

Then in your reasoning, you can't assume $0 < b$. It looks like you have currently assumed that here:

If $b \leq a_n \forall a_n\in \mathbb{R}$

then $0 < b \leq a_n < b_n$

But if you replace this with $b \le a_n < b_n$ (just remove "$0 < $"), and apply the same idea, you should get the conclusion you want.